How do you find a formula for the sum to #n# terms of the sequence #1^4, 2^4, 3^4, 4^4, 5^4, 6^4,...# ?
1 Answer
Explanation:
The formula for the sum to
Such a polynomial will be determined by
#color(blue)(1), 17, 98, 354, 979, 2275#
Write down the sequence of differences between consecutive terms:
#color(blue)(16), 81, 256, 625, 1296#
Write down the sequence of differences of those differences:
#color(blue)(65), 175, 369, 671#
Write down the sequence of differences of those differences:
#color(blue)(110), 194, 302#
Write down the sequence of differences of those differences:
#color(blue)(84), 108#
Write down the sequence of differences of those differences:
#color(blue)(24)#
Having taken the difference
#s_n = color(blue)(1)/(0!)+color(blue)(16)/(1!)(n-1)+color(blue)(65)/(2!)(n-1)(n-2)+color(blue)(110)/(3!)(n-1)(n-2)(n-3)+color(blue)(84)/(4!)(n-1)(n-2)(n-3)(n-4)+color(blue)(24)/(5!)(n-1)(n-2)(n-3)(n-4)(n-5)#
#color(white)(s_n) = 1+16(n-1)+65/2(n^2-3n+2)+55/3(n^3-6n^2+11n-6)+7/2(n^4-10n^3+35n^2-50n+24)+1/5(n^5-15n^4+85n^3-225n^2+274n-120)#
#color(white)(s_n) = 1/30 n(6n^4+15n^3+10n^2-1)#
This formula is guaranteed to match the six sample values we initially listed, so is the correct quintic.